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The solution of the differential equation xdydx=2x2+1x\frac{dy}{dx} = 2x^2 + 1; x≠0x \neq 0 with y=1y = 1 when x=1x = 1, is

Solution

✅ Correct Option: 4

dydx=2x+1x\frac{dy}{dx} = 2x+\frac{1}{x}, so y=x2+log⁡e∣x∣+Cy = x^2+\log_e|x|+C.

Using y=1y=1 at x=1x=1: 1=1+0+C⇒C=01 = 1+0+C \Rightarrow C=0.

Hence y=x2+log⁡e∣x∣y = x^2+\log_e|x|.

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